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kolbaska11 [484]
3 years ago
12

Solve: x^+4/x-1 = 5/x-1

Mathematics
1 answer:
AlexFokin [52]3 years ago
7 0
The first step for solving this equation is to determine the defined range.
\frac{ x^{4}  }{x-1} = \frac{5}{x-1}, x ≠ 1
Remember that when the denominators of both fractions are the same,, you need to set the numerators equal. This will look like the following:
x^{4} = 5
Take the root of both sides of the equation and remember to use both positive and negative roots.
x +/- \sqrt[4]{5}
Separate the solutions.
x = \sqrt[4]{5}           , x ≠ 1
x = -\sqrt[4]{5} 
Check if the solution is in the defined range.
x = \sqrt[4]{5} 
x = -\sqrt[4]{5} 
This means that the final solution to your question are the following:
x = \sqrt[4]{5}          
x = -\sqrt[4]{5} 
Let me know if you have any further questions.
:)
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If w=10, x=5, and y=6, what is the surface are of the figure?
Olin [163]

Answer:

hello!!!!

the answer is 280

im not really sure... read the description.... since i dont see a picture attached i dont know the figure shape which makes it difficult to guarantee the correct answer for you....

please read the explanation.. it always helps...

Step-by-step explanation:

okay im going to guess that the figure is a rectangular prism... if not please correct me

if so.... then...

2(10*5 + 10*6 + 5*6)= 280

7 0
2 years ago
Help me put this in order please help thanks
KATRIN_1 [288]
1. 8
2. 15
3. 2
4. 1/2
5. 3
those are the answers in order, hope this helps!
5 0
3 years ago
Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

So, you have proved the initial statement

4 0
3 years ago
What is the standard form of the equation y=-5/2x-3 ?
butalik [34]

The standard form of an equation is the form where that equation has no fractions and is written in the form ax + by = c.


To get rid of the fractions in y = -5/2 x - 3, we need to multiply both sides by 2. That makes the equation become:


2y = -5x - 6


Now we need to get it into the form ax + by = c. We can do this by adding 5x to each side.


5x + 2y = -6 is our answer

6 0
3 years ago
Solve the simultaneous equations​
nexus9112 [7]

Answer:

Step-by-step explanation:

2x + 3y = 2

x = 2 - 3y /2

x + y = 0

substitute the value of x

\frac{2 \\- 3y}{2} + y = 0

2 - 3y + 2y = 0

2 = y

again,substitute the value of y

x = 2 - 3y / 2

=2 - 3*2 / 2

=2 - 6 / 2

=-2

6 0
2 years ago
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